Peak Current

Let us expand on Steven's coding solution here. The basic attack mode is to simplify our passive components into a single impedance value, $Z_{eq}.$ The circuit has one branch with a resistor in series with an inductor, and the other parallel branch has a resistor in series with a capacitor. Knowing that $\omega = 100$ Hz from the AC voltage source, we calculate $Z_{eq}$ according to:

$\Large \frac{1}{Z_{eq}} = \frac{1}{R_{1}+Z_{L}} + \frac{1}{R_{2}+Z_{C}} = \frac{1}{R_{1}+j\omega L} + \frac{1}{R_{2}+1/j\omega C}$ (where $j=\sqrt{-1}$ );

or $\Large \frac{1}{Z_{eq}} = \frac{1}{10+j(100)(0.1)} + \frac{1}{10-j/(100)(0.001)}$ ;

or $\Large \frac{1}{Z_{eq}} =\frac{1}{10+10j} + \frac{1}{10 - 10j}$ ;

or $\Large \frac{1}{Z_{eq}} =\frac{(10-10j)+(10+10j)}{(10+10j)(10-10j)};$

or $\Large \frac{1}{Z_{eq}} =\frac{20}{100+100};$

or $\Large \frac{1}{Z_{eq}} =\frac{1}{10};$

or $\large Z_{eq} = 10\Omega.$

By Ohm's Law, the current computes to:

$\large I(t) = \frac{E(t)}{Z_{eq}} = \frac{50\sin(100t)}{10} = 5\sin(100t) \Rightarrow I_{MAX} = \boxed{5A}.$

Since no initial conditions are given, this must be an AC steady state problem. Calculations and comments are given in the attached code:

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import math

Epeak = 50.0           # voltage source peak
w = 100.0              # angular frequency

R1 = 10.0              # resistor values
R2 = 10.0

C = 0.001           # L and C values
L = 0.1

XL = w*L            # L and C reactances and complex impedances
XC = 1.0/(w*C)

ZL = complex(0.0,XL)
ZC = complex(0.0,-XC)

Z1 = R1 + ZL        # parallel branch impedances
Z2 = R2 + ZC

Zp = Z1*Z2/(Z1+Z2)  # total equivalent impedance

Ipeak = Epeak / abs(Zp)  # current peak

print Ipeak
#5.0